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For a compound event $E_1E_2$,
Pr($E_1E_2$) = Pr($E_1$)Pr($E_2$|$E_1$)
such that if $E_1$ and $E_2$ are independent events, we can say that:
Pr($E_1E_2$) = Pr($E_1$)Pr($E_2$)

Is there an analogous formula that can be derived for the conditional probability: Pr($E_1E_2$|$E_3$)?

I would like to be able to reduce Pr($E_1E_2$|$E_3$) to terms that do not involve all three of the events $E_1$, $E_2$, $E_3$, given the assumption that only $E_1$ and $E_3$ are independent.

If this cannot be done, then what assumptions would be necessary in order to be able to do it?

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2 Answers 2

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Here is one way to write it given the assumption: $$ P(E_1\cap E_2\mid E_3)=P(E_2\mid E_1\cap E_3)P(E_1\mid E_3)=P(E_2\mid E_1\cap E_3)P(E_1). $$

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  • $\begingroup$ Is there no way to re-write it as a product of terms, each of which only involves one or two out of the three events? $\endgroup$
    – user1247
    Commented Jan 28, 2013 at 10:40
  • $\begingroup$ I don't think there is, but I am not sure. You should probably edit your question and state you're looking for such an expression. $\endgroup$ Commented Jan 28, 2013 at 10:45
  • $\begingroup$ OK I edited my question. I guess this boils down to something like: if A and B are independent, can I write P(A|BC)=P(A|C)? If not, what are the minimal assumptions that are necessary in order for that to be true? $\endgroup$
    – user1247
    Commented Jan 28, 2013 at 10:55
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Hint: If you write out $$Pr(E2|E1)=\frac{Pr(E_2\cap E_1)}{Pr(E_1)}$$ then you can investigate more complicated formulas involving more events. Don't remember a formula but understand them.

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  • $\begingroup$ The problem is that I can't figure out how using that formula would enable me to show, for example, that P(A|BC)=P(A|C), given A and B being independent. $\endgroup$
    – user1247
    Commented Jan 28, 2013 at 12:32

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